class Solution:
  def maximumXOR(self, nums: list[int]) -> int:
    # 1. nums[i] & (nums[i] ^ x) enables you to turn 1-bit to 0-bit from
    #    nums[i] since x is arbitrary.
    # 2. The i-th bit of the XOR of all the elements is 1 if the i-th bit is 1
    #    for an odd number of elements.
    # 3. Therefore, the question is equivalent to: if you can convert any digit
    #    from 1 to 0 for any number, what is the maximum for XOR(nums[i]).
    # 4. The maximum we can get is of course to make every digit of the answer
    #    to be 1 if possible
    # 5. Therefore, OR(nums[i]) is an approach.
    return functools.reduce(operator.ior, nums)
